Question

Hydrogen at normal condition 101,3 kPa(a) and 25 C degree.

Please calculate volumetric energy density (for LHV)
Answer give in MJ/nm3 (normalcubicmeter). MJ = mega Joule=10^6 Joule

Answer 1

To calculate the volumetric energy density for hydrogen at normal conditions (101.3 kPa(a) and 25 degrees Celsius), we need to use the Lower Heating Value (LHV) of hydrogen.

The LHV of hydrogen is approximately 120 MJ/kg. We can convert this value to MJ/nm³ by considering the density of hydrogen at normal conditions. The density of hydrogen at standard temperature and pressure (STP) is approximately 0.08988 kg/m³.

To find the volumetric energy density (V) in MJ/nm³, we can use the following formula:

V = (LHV * density) / 1000

Substituting the values:

V = (120 MJ/kg * 0.08988 kg/m³) / 1000
V ≈ 0.01077 MJ/nm³

Therefore, the volumetric energy density of hydrogen at normal conditions is approximately 0.01077 MJ/nm³ (megajoules per normal cubic meter).